The number of 7-digit numbers which are multiples of 1 and are formed using all the digits 1, 2, 3, 4, 5, 7 and 9 is
The number of 7-digit numbers which are multiples of 1 and are formed using all the digits 1, 2, 3, 4, 5, 7 and 9 is
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1 Answer
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Sum of all given numbers = 31
Difference between odd and even positions must be 0,11 or 22 but 0 and 22 are not possible.
Hence 11 is possible.
This is possible only when either 1, 2, 3, 4 if filled in odd places in order and remaining in other order.
Hence 2, 3, 5 or 7, 2, 1 or 4, 5, 1 at even places.
Total possible ways = (4! × 3!) × 4 = 576
Similar Questions for you
Start with
(1)
(2)
(3) GTE : 4!, GTN: 4!, GTT : 4!
(4) GTWENTY = 1
⇒ 360 + 60 + 60 + 24 + 24 + 24 + 1 = 553
x + 2y + 3z = 42
0 x + 2y = 42 ->22 cases
1 x + 2y = 39 ->19 cases
2 x + 2y = 36 ->19 cases
3 x + 2y = 33 ->17 cases
4 x + 2y = 30 ->16 cases
5 x + 2y = 27 ->14 cases
6 x + 2y = 24 ->13 cases
7 x + 2y = 21 ->11 cases
8 x + 2y = 18 ->10 cases
9 x + 2y = 15 ->8 cases
10 x + 2y =12 -> 7 cases
11 x + 2y = 9 -> 5 cases
12 x + 2y = 6 -> 4 cases
13 x + 2y = 3 -> 2 cases
14 x + 2y = 0 -> 1 cases.
Total ways to partition 5 into 4 parts are:
5 0
4 1 0
3 2 0
3 1 0
2 1
51 Total way
After giving 2 apples to each child 15 apples left now 15 apples can be distributed in
15+3–1C2 = 17C2 ways
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